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Im studying for my exam in linear algebra in a few weeks. Can someone explain this a bit better to me. I have solved a matrice $ T \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ \end{bmatrix}$ = $ \begin{bmatrix} x_{1} - x_{2} + x_{3} - x_{4}\\ x_{1} + x_{2} + 3x_{3} + 3x_{4}\\ 2x_{1} + x_{2} + 5x_{3} + 4x_{4}\\ 3x_{1} + 2x_{2} + 8x_{3} + 7x_{4}\\ \end{bmatrix} $

I have almost solved it. I get to the fact that this can be simplified to 4 unknowns but only 2 indepedent equations and 2 free parameters. The assignment is finding a basis for kernel and range of linear transformations given by the system above. I understand this the whole way til the final end. (This solution is already given to me and I need some help understanding the last part). From what I know free variables have a corresponding vector. Theese vectors form the basis of a x-dimensional space.

The two independent equations are now after solving the system: $$ x_1 - x_2 + x_3 - x_4 = 0 $$ $$ x_2 + x_3 + 2x_4 = 0 $$

I set $$ x_4 = s$$ $$x_3 = t $$

then; $$ x_2 = -2x_4 - x_3 = -2s-t$$ $$ x_1 = x_2 - x_3 + x_4 = -s-2t $$

So far I understand but here comes the part I am missing; this tells me according to the solutions that $$ \overline{x} \in Ker(T) \leftrightarrow \overline{x} = s \begin{bmatrix} -1\\ -2\\ 0\\ 1\\ \end{bmatrix} + t \begin{bmatrix} -2\\ -1\\ 1\\ 0\\ \end{bmatrix} $$

And I just don't understand the last part, where did the vectors come from ? Did I count s's and t's from the last matrix solution, which was: $$\begin{bmatrix} 1&-1&1&-1\\ 0& 1& 1&2\\ 0&0&0&0\\ 0&0&0&0\\ \end{bmatrix}$$

So my question is, where does the vectors t and s come from?

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  • $\begingroup$ I believe you are confusing yourself, you kind of mentioned it yourself that you set $x_4=s, x_3=t$ where $s,t \in \mathbb{R}$ (I suppose) and therefore these aren't vectors but scalar identities. If you replug them into your remaining equations and then factor out, you will obtain the solution. $\endgroup$ – Spaced Oct 10 '15 at 13:13
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I assumed you asked about the vectors $$\left[\begin{array}{rr}-1\\-2\\0\\1\end{array}\right]\text{ and } \left[\begin{array}{rr}-2\\-1\\1\\0\end{array}\right].$$ The reason here is that $$ x_2 = -2x_4 - x_3 = -2s-t$$ $$ x_1 = x_2 - x_3 + x_4 = -s-2t.$$ So $$\bar x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}-2s-t\\-s-2t\\0+t\\s+0\end{bmatrix},$$ separating $s$ and $t$ (as they are independent scalars): $$\bar x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}-2s\\-s\\0\\s\end{bmatrix}+\begin{bmatrix}-t\\-2t\\t\\0\end{bmatrix}.$$ Putting $s$ and $t$ outside of each vector above, you would get the final linear combination.

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  • $\begingroup$ I think you wrote a bit wrong the x_1 and x_2 but I was actually just confusing myself, and it helped just reading yours and doing it again. I totally see it now. I have my s and t already when i plug it in with the remaining eq and i did that before too and made a mistake so thats why I started confusing myself. Thanx anyway :) $\endgroup$ – Elin Hägglund Oct 10 '15 at 16:17
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When you write $x=(x_1,x_2,x_3,x_4,x_5)$, $x$ is the vector while $x_1,x_2,x_3,x_4,x_5$ are scalars from the field ,(say $\mathbb{R}$).

So if you write $x_3=t;$ and $x_4=s$, it means $s,t$ are scalars.

Since $x_1, x_2$ depend on $x_3,x_4$, by writing $x_3=t;$ and $x_4=s$, you are allowing $x_3, x_4$ to vary over $\mathbb{R}$ and depending on it you will get the $\overline{x} $.

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As you have an underdetermined equation system we need two scalars $s$ and t to define parameters for the space of solutions.

The vectors to describe the solutions depend on the choice of $s$ and $t$ relative to $x_k$.

You can think of it as picking a coordinate system on a plane piece of paper. You can have many different axes, $s$ and $t$ would be the coordinates for your paper and they have some relation to the other coordinate system with $x$s in it. Although in this example the solution space is 2D plane and the original space is 3D instead of 4D but the same ideas apply.


Oh wait you also seem a bit confused on how to translate the parameterization to solutions in terms of $T$. What you have calculated is just a way to represent the rows in $T$. I.e. each row should be possible to write as a linear combination in the vectors in $\ker(T)$. The vectors you have found (next to $s$ and $t$) span the space that $T$ misses and the first two rows in the matrix span what it can measure.

$$T = \left[\begin{array}{cc}1&0\\1&2\\2&3\\3&5 \end{array}\right]\left[\begin{array}{rrrr} 1&-1&1&-1\\0&1&1&2\end{array}\right] $$

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